How to solve 2/x+3=3/5-x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2/x+3=3/5-x | Equatix

Question

$$2/x+3=3/5-x$$

Solve for x (complex solution)

$x=\frac{-6+\sqrt{14}i}{5}\approx -1.2+0.748331477i$
$x=\frac{-\sqrt{14}i-6}{5}\approx -1.2-0.748331477i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $5x$, the least common multiple of $x,5$.

$$5\times 2+5x\times 3=5x\times \frac{3}{5}-x\times 5x$$

Multiply $5$ and $2$ to get $10$.

$$10+5x\times 3=5x\times \frac{3}{5}-x\times 5x$$

Multiply $5$ and $3$ to get $15$.

$$10+15x=5x\times \frac{3}{5}-x\times 5x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$10+15x=5x\times \frac{3}{5}-x^{2}\times 5$$

Cancel out $5$ and $5$.

$$10+15x=3x-x^{2}\times 5$$

Multiply $-1$ and $5$ to get $-5$.

$$10+15x=3x-5x^{2}$$

Subtract $3x$ from both sides.

$$10+15x-3x=-5x^{2}$$

Combine $15x$ and $-3x$ to get $12x$.

$$10+12x=-5x^{2}$$

Add $5x^{2}$ to both sides.

$$10+12x+5x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$5x^{2}+12x+10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $12$ for $b$, and $10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\times 10}}{2\times 5}$$

Square $12$.

$$x=\frac{-12±\sqrt{144-4\times 5\times 10}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-12±\sqrt{144-20\times 10}}{2\times 5}$$

Multiply $-20$ times $10$.

$$x=\frac{-12±\sqrt{144-200}}{2\times 5}$$

Add $144$ to $-200$.

$$x=\frac{-12±\sqrt{-56}}{2\times 5}$$

Take the square root of $-56$.

$$x=\frac{-12±2\sqrt{14}i}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{-12±2\sqrt{14}i}{10}$$

Now solve the equation $x=\frac{-12±2\sqrt{14}i}{10}$ when $±$ is plus. Add $-12$ to $2i\sqrt{14}$.

$$x=\frac{-12+2\sqrt{14}i}{10}$$

Divide $-12+2i\sqrt{14}$ by $10$.

$$x=\frac{-6+\sqrt{14}i}{5}$$

Now solve the equation $x=\frac{-12±2\sqrt{14}i}{10}$ when $±$ is minus. Subtract $2i\sqrt{14}$ from $-12$.

$$x=\frac{-2\sqrt{14}i-12}{10}$$

Divide $-12-2i\sqrt{14}$ by $10$.

$$x=\frac{-\sqrt{14}i-6}{5}$$

The equation is now solved.

$$x=\frac{-6+\sqrt{14}i}{5}$$ $$x=\frac{-\sqrt{14}i-6}{5}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $5x$, the least common multiple of $x,5$.

$$5\times 2+5x\times 3=5x\times \frac{3}{5}-x\times 5x$$

Multiply $5$ and $2$ to get $10$.

$$10+5x\times 3=5x\times \frac{3}{5}-x\times 5x$$

Multiply $5$ and $3$ to get $15$.

$$10+15x=5x\times \frac{3}{5}-x\times 5x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$10+15x=5x\times \frac{3}{5}-x^{2}\times 5$$

Cancel out $5$ and $5$.

$$10+15x=3x-x^{2}\times 5$$

Multiply $-1$ and $5$ to get $-5$.

$$10+15x=3x-5x^{2}$$

Subtract $3x$ from both sides.

$$10+15x-3x=-5x^{2}$$

Combine $15x$ and $-3x$ to get $12x$.

$$10+12x=-5x^{2}$$

Add $5x^{2}$ to both sides.

$$10+12x+5x^{2}=0$$

Subtract $10$ from both sides. Anything subtracted from zero gives its negation.

$$12x+5x^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$5x^{2}+12x=-10$$

Divide both sides by $5$.

$$\frac{5x^{2}+12x}{5}=-\frac{10}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\frac{12}{5}x=-\frac{10}{5}$$

Divide $-10$ by $5$.

$$x^{2}+\frac{12}{5}x=-2$$

Divide $\frac{12}{5}$, the coefficient of the $x$ term, by $2$ to get $\frac{6}{5}$. Then add the square of $\frac{6}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=-2+\left(\frac{6}{5}\right)^{2}$$

Square $\frac{6}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=-2+\frac{36}{25}$$

Add $-2$ to $\frac{36}{25}$.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=-\frac{14}{25}$$

Factor $x^{2}+\frac{12}{5}x+\frac{36}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{6}{5}\right)^{2}=-\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{-\frac{14}{25}}$$

Simplify.

$$x+\frac{6}{5}=\frac{\sqrt{14}i}{5}$$ $$x+\frac{6}{5}=-\frac{\sqrt{14}i}{5}$$

Subtract $\frac{6}{5}$ from both sides of the equation.

$$x=\frac{-6+\sqrt{14}i}{5}$$ $$x=\frac{-\sqrt{14}i-6}{5}$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Solve for x (complex solution)

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square